Trying to get 2018 (2)

Logic Level 3

I have five kinds of cards(there are unlimited amount of them), 1 \large\boxed{ 1 } , + \large\boxed{ + } , × \large\boxed{\times} , ( \large\boxed{(} , ) \large\boxed{)} .

I can get numbers by putting the cards in a row (no exponential),and I can put several 1 \large\boxed{ 1 } s together.

For example,I can get to 14 14 by ( 1 + 1 ) ( ( 1 + 1 ) ( 1 + 1 + 1 ) + 1 ) (1+1)((1+1)(1+1+1)+1) or 11 + 1 + 1 + 1 11+1+1+1 .

I want to use these cards to get to 2018 2018 ,what is the least amount of 1 \large\boxed{ 1 } s I have to use?

See similar problem here


The answer is 13.

X X by X X , 17, Taiwan

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1 solution

X X
May 13, 2018

2018 = 2 × 1009 = 2 ( 9 × 112 + 1 ) = ( 1 + 1 ) [ ( 111 + 1 ) ( 1 + 1 + 1 ) ( 1 + 1 + 1 ) + 1 ] 2018=2\times1009=2(9\times112+1)=(1+1)[(111+1)(1+1+1)(1+1+1)+1]

0 Helpful 0 Interesting 0 Brilliant 0 Confused

You can make it twelve: ( 1 + 1 ) × ( ( 111 + 1 ) × ( 1 + 1 + 1 ) ( 1 + 1 ) ) (1+1)\times \left((111+1)\times(1+1+1)^{(1+1)}\right)

Blan Morrison - 3 years ago

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I editted the problem.Thanks for pointing out.

X X - 3 years ago

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Is it okay if I post a part three with this as the solution?

Blan Morrison - 3 years ago

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@Blan Morrison – It's okay,I don't mind.

X X - 3 years ago

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